Ring of integers of $f(X)=X^3+X^2-2X+8$ is principal Let $K=\Bbb Q[x]$ be a cubic number field with $x$ be a root of $f(X)=X^3+X^2-2X+8$. I want to show that $A$, the ring of integers of $K$ is principal.
What I have shown is that $A=\Bbb Z+\Bbb Zx+\Bbb Zy$ where $y=4/x$(edited) and $(2)=(2,x,y)(2,x+1,y)(2,x,y+1)$ is the product of distinct ideals of norm 2 (by brutal method), and Minkowski bound is more than $6$, so it suffices to show that every prime factor $(2),(3),(5)$ whose norm $\le 6$ is principal.
But I don't know how to proceed. The hint says that $(x-2)/(x+2), (x-1)/(x+3)$ are the generators of two of the three prime ideals of norm 2, and $(3)$ is prime, and $(5)$ is a product of prime ideal of norm 5 and norm 25, and compare $(5)$, $(x+1)$. How can I proceed?
 A: Let $\alpha$ be a root of $f(x)=x^3+x^2-2x+8$. Are you sure that $1$, $\alpha$, $\frac{1}{4}\alpha$ is an integral basis? You might want to take a look at the minimal polynomial of $\frac{1}{4}\alpha$.
An integral basis is actually given by $1$, $\alpha$, $\beta=\frac{1}{2}(\alpha+\alpha^2)$. You then factor the ideals $(2)$, $(3)$, and $(5)$.


*

*Let $\mathfrak{p}_2=(2,\beta)$, and $\mathfrak{q}_2=(2,\beta+1)$. Then $(2)=\mathfrak{p}_2\mathfrak{q}_2^2$.

*$(3)$ is inert.

*Let $\mathfrak{p}_5=(5,\alpha+1)$, and $\mathfrak{p}_{25}=(5,\alpha^2+3)$. Then $(5)=\mathfrak{p}_5\mathfrak{p}_{25}$.


The hint then seems to have done most of the hard work for you. You've already said that it's given you $\frac{\alpha-2}{\alpha+2}$ as a generator for one of the prime ideals of norm $2$, and $\frac{\alpha-1}{\alpha+3}$ as a generator of the other.
You also say to look at $(\alpha+1)$, and $\text{Nm}(\alpha+1)=10=2\cdot 5$, so you instantly find two elements of norm $\pm 5$, namely
$$
(\alpha+1)\frac{\alpha+2}{\alpha-2}\text{ and }(\alpha+1)\frac{\alpha+3}{\alpha-1}
$$
Can you show that one of these generates $\mathfrak{p}_5$?
